Total electric charge on the earth,

In summary: I was too lazy! In summary, the Earth has a total charge of pi rho_s R^3. Gauss's law can be used to find an expression for E at the point r outside the Earth (r>R). E within the Earth's interior (r<R) can also be found using gauss's law.
  • #1
vorcil
398
0

Homework Statement


The Earth can be regarded as a sphere of radius R,
the volume density of charge distributed as [tex] \rho = \frac{\rho_s r}{R} [/tex]
where the density is 0 at thcce centre and rises linearly with radius until it reaches ps at the Earth's surface

i) prove that the total charge on the Earth is [tex] \pi \rho_s R^3 [/tex]

ii) use gauss's law to find an expression for E at the point r outside the Earth (r>R)

iii) use gauss's law to find an expression for E within the Earth's interior (r<R)

iv) write expressions for the x-, y- and z- components of E at a point outside the Earth and calculate [tex] \bf{\nabla .E }

Homework Equations



Qenclosed = [tex] \int_v \rho d\tau. [/tex] (for a volume)

[tex] \oint_S \bf{E} . d\bf{a} = \frac{1}{\epsilon} Qenclosed [/tex]

The Attempt at a Solution



[tex] \int_v \rho \dtau [/tex]

substituting in the given rho,

[tex] \int \rho_s \frac{r}{R} \dtau[/tex]

the outer radius r is = R so it just becomes R/R =1,

[tex] \rho_s *1 \int \dtau [/tex]

this is where I integrate the dtau and it should give me the volume of a sphere

[tex] \rho_s \frac{4}{3}\pi r^3 [/tex]

this sort of looks like the answer I'm looking for
HOWEVER I DONT KNOW HOW TO GET RID OF THE 4/3 constant!
what have I done wrong?
i've tried heaps of other methods, but I'm always getting some type of constant out the front
 
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  • #2
The density is proportional to r, as you wrote, but you can't pull this factor of r outside the integral. You should write:
[tex]\int_0^R\rho(r)d\tau = \int_0^R\rho_s\frac{r}{R}4\pi r^2 dr = 4\pi\frac{\rho_s}{R}\int_0^Rr^3 dr = 4\pi \frac{\rho_s}{R} \frac{R^4}{4} = \pi\rho_sR^3[/tex]

Does this make sense?
 
  • #3
phyzguy said:
The density is proportional to r, as you wrote, but you can't pull this factor of r outside the integral. You should write:
[tex]\int_0^R\rho(r)d\tau = \int_0^R\rho_s\frac{r}{R}4\pi r^2 dr = 4\pi\frac{\rho_s}{R}\int_0^Rr^3 dr = 4\pi \frac{\rho_s}{R} \frac{R^4}{4} = \pi\rho_sR^3[/tex]

Does this make sense?

Yes thanks, I actually ended up doing it like this myself last night, I just didn't update it here on PF
 

What is the total electric charge on the earth?

The total electric charge on the earth is constantly changing due to the movement of charged particles in the atmosphere and the earth's interior. However, it is estimated to be around 500,000 Coulombs at any given time.

How is the total electric charge on the earth measured?

The total electric charge on the earth is measured using a device called an electrostatic voltmeter. This instrument measures the electric potential difference between the earth and the surrounding air, which is then used to calculate the total electric charge.

What factors contribute to the total electric charge on the earth?

The total electric charge on the earth is influenced by various factors such as thunderstorms, solar activity, and the earth's rotation. These factors can cause an accumulation or depletion of electric charge in certain regions of the earth.

Can the total electric charge on the earth be negative?

Yes, the total electric charge on the earth can be negative. This is typically seen in regions of the earth where there is a buildup of negative ions, such as during thunderstorms or in areas with high levels of air pollution.

What impact does the total electric charge on the earth have on our daily lives?

The total electric charge on the earth can have various effects on our daily lives, such as influencing the weather, creating static electricity, and affecting the operation of electronic devices. It can also play a role in the formation of lightning and auroras.

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